For a suitably chosen real constant $$a$$, let a function, $$f : \mathbb{R} - \{-a\} \to \mathbb{R}$$ be defined by $$f(x) = \frac{a-x}{a+x}$$. Further suppose that for any real number $$x \neq -a$$, and $$f(x) \neq -a$$, $$(f \circ f)(x) = x$$. Then $$f\left(-\frac{1}{2}\right)$$ is equal to:

First we recall that the composite function $$(f \circ f)(x)$$ means “apply $$f$$ twice”, i.e. $$(f \circ f)(x)=f\bigl(f(x)\bigr).$$ The statement of the question says that this composite equals $$x$$ itself. Symbolically, for every real $$x \neq -a$$ with $$f(x)\neq -a,$$ we have the requirement

$$f\bigl(f(x)\bigr)=x.$$

Such a function is called an involution. We shall use this condition to determine the constant $$a$$ and then evaluate $$f\!\left(-\dfrac12\right).$$

The given function is

$$f(x)=\dfrac{a-x}{\,a+x\,}.$$

We first compute $$y=f(x):$$

$$y=\dfrac{a-x}{a+x}.$$

Now we apply $$f$$ again, replacing its argument by $$y$$. Thus

$$f\bigl(f(x)\bigr)=\dfrac{a-y}{\,a+y\,}.$$

Substituting $$y=\dfrac{a-x}{a+x}$$ we obtain

$$f\bigl(f(x)\bigr)=\dfrac{a-\dfrac{a-x}{a+x}}{\,a+\dfrac{a-x}{a+x}\,}.$$

To simplify, we bring every term over the common denominator $$(a+x).$$ For the numerator:

$$a-\dfrac{a-x}{a+x}=\dfrac{a(a+x)-(a-x)}{a+x}=\dfrac{a^2+ax-a+x}{a+x}.$$

Collecting like terms gives

$$a^2+ax-a+x=a^2-a+x(a+1).$$

So the numerator equals

$$\dfrac{a^2-a+x(a+1)}{a+x}.$$

For the denominator:

$$a+\dfrac{a-x}{a+x}=\dfrac{a(a+x)+(a-x)}{a+x}=\dfrac{a^2+ax+a-x}{a+x}.$$

Collecting like terms gives

$$a^2+ax+a-x=a^2+a+x(a-1).$$

Hence the denominator equals

$$\dfrac{a^2+a+x(a-1)}{a+x}.$$

Therefore

$$f\bigl(f(x)\bigr)=\dfrac{\dfrac{a^2-a+x(a+1)}{a+x}}{\dfrac{a^2+a+x(a-1)}{a+x}}=\dfrac{a^2-a+x(a+1)}{a^2+a+x(a-1)}.$$

The requirement $$f\bigl(f(x)\bigr)=x$$ now becomes the identity

$$\dfrac{a^2-a+x(a+1)}{a^2+a+x(a-1)}=x \quad \text{for all admissible } x.$$

Cross-multiplying (which is valid because denominators are non-zero away from the excluded points) we get

$$a^2-a+x(a+1)=x\bigl[a^2+a+x(a-1)\bigr].$$

Expanding the right-hand side yields

$$a^2-a+x(a+1)=a^2x+ax+(a-1)x^2.$$

Now we bring every term to the right so that the left side becomes zero, collecting like powers of $$x$$:

$$0=a^2x+ax+(a-1)x^2-\bigl(a^2-a+ax+x\bigr).$$

Distributing the negative sign and combining terms one by one, we obtain

$$0=\underbrace{(a-1)}_{\text{coefficient of }x^2}x^2+\underbrace{\bigl(a^2-1\bigr)}_{\text{coefficient of }x}x+\underbrace{\bigl(-a(a-1)\bigr)}_{\text{constant term}}.$$

For the polynomial to be identically zero for all real $$x,$$ each individual coefficient must vanish:

$$a-1=0,\qquad a^2-1=0,\qquad -a(a-1)=0.$$

The first of these directly gives $$a=1.$$ Substituting $$a=1$$ into the second and third equations confirms that they are also satisfied. Thus the only admissible constant is

$$a=1.$$

With this value, the original function becomes

$$f(x)=\dfrac{1-x}{\,1+x\,}.$$

We are required to evaluate $$f\!\left(-\dfrac12\right).$$ Substituting $$x=-\dfrac12$$ gives

$$f\!\left(-\dfrac12\right)=\dfrac{1-\left(-\dfrac12\right)}{1+\left(-\dfrac12\right)}=\dfrac{1+\dfrac12}{1-\dfrac12}=\dfrac{\dfrac32}{\dfrac12}.$$

Dividing the two fractions, we obtain

$$f\!\left(-\dfrac12\right)=\dfrac32 \times \dfrac21=3.$$

Hence, the correct answer is Option D.